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प्रश्न
An electron moving horizontally with a velocity of 4 ✕ 104 m/s enters a region of uniform magnetic field of 10−5 T acting vertically upward as shown in the figure. Draw its trajectory and find out the time it takes to come out of the region of magnetic
field.
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उत्तर
Let the time taken by the electron to come out of the region of magnetic field be t.
Velocity of the electron, v = 4 × 104 m/s
Magnetic field, B = 10−5 T
Mass of the electron, m = 9 × 10−31 kg
We know
\[t = \frac{\pi r}{v}\]
\[\text { where r } = \frac{mv}{Bq}\]
\[\text { Now,} \]
\[t = \frac{\pi m}{Bq} = \frac{3 . 14 \times 9 \times {10}^{- 31}}{{10}^{- 5} \times 1 . 6 \times {10}^{- 19}}\]
\[ \Rightarrow t = 17 . 66 \times {10}^{- 7} s = 1 . 76 \mu s\]
Thus, the time taken by the electron to come out of the region of magnetic field is 1.76 μs.
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